a. Give and explain the decreasing order of atomic radius of fluorine `(F)`, nitrogen `(N)` and oxygen `(O)`.
b. Give the decreasing order of van der Waals radii: `N, O, H, Cl, Br`.

Explanation

a. Decreasing order of atomic radius is `O gt F gt N`.
Explanation: The atomic size generally decreases across a period as shown for the elements of the second period. Because within period the outer electrons are in the same valence shell and the effective nuclear charge `(Z_(eff))` increases as the atomic number increases resulting in the increased attraction of electrons to the nucleus.
So, the decreasing order of atomic radius should be `N gt O gt F`.
But it is not observed, since from `N` to `O, Z_(eff))` increases, so the size should decrease from `N` to `O`. But in oxygen pairing of electrons takes place, so due to the repulsion between paired electrons, size of `O gt N`. `[N = 2s^(2) 2p^(3) , O = 2s^(2) 2p^(4)]`
Similarly, due to the repulsion between two lone pairs of electrons in `F (2s^(2) 2p^(5))`, the size should increase from `O` to `F`. But high nuclear charge of `F` than that of `O`, compensates the repulsion between two pairs of electrons. Hence, size of `O gt F`.
Thus decreasing order of atomic radius is `O gt F gt N`.
b. `r_(vdw) prop` Numbers of energy shells and
`r_(vdw) prop (1)/("Nuclear charge")`
i. `H` has only one energy shell `(n = 1)`, `Cl` has three `(n = 3)` and `Br` has four `(n = 4)`. Therefore,
`r_(vdw)` of `Br gt r_(vdw)` og `Cl gt r_(vdw)` of `H`
ii. Both `N` and `O` have two energy shells `(n = 2)`, but `Z` on oxygen `(+8)` is greater than on `N (+7)`. Therefore,
`r_(vdw)` of `N gt r_(vdw)` of `O`
Hence, the decreasing order of `r_(vdw)` is
`Br gt Cl gt N gtO gt H`